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  • POJ 1811 Prime Test

    Prime Test

    Time Limit: 6000ms
    Case Time Limit: 4000ms
    Memory Limit: 65536KB
    This problem will be judged on PKU. Original ID: 1811
    64-bit integer IO format: %lld      Java class name: Main
     
    Given a big integer number, you are required to find out whether it's a prime number.
     

    Input

    The first line contains the number of test cases T (1 <= T <= 20 ), then the following T lines each contains an integer number N (2 <= N < 254).
     

    Output

    For each test case, if N is a prime number, output a line containing the word "Prime", otherwise, output a line containing the smallest prime factor of N.
     

    Sample Input

    2
    5
    10
    

    Sample Output

    Prime
    2
    

    Source

     
    解题:Miller-Rabbin + Pollard-rho
     
     1 #include <cstdio>
     2 #include <ctime>
     3 #include <cstdlib>
     4 using namespace std;
     5 typedef long long LL;
     6 const int maxn = 1001;
     7 LL mul(LL a,LL b,LL mod) {
     8     if(!a) return 0;
     9     return ((a&1)*b%mod + (mul(a>>1,b,mod)<<1)%mod)%mod;
    10 }
    11 LL quickPow(LL a,LL d,LL n) {
    12     LL ret = 1;
    13     while(d) {
    14         if(d&1) ret = mul(ret,a,n);
    15         d >>= 1;
    16         a = mul(a,a,n);
    17     }
    18     return ret;
    19 }
    20 bool check(LL a,LL d,LL n) {
    21     if(n == a) return true;
    22     while(~d&1) d >>= 1;
    23     LL t = quickPow(a,d,n);
    24     while(d < n-1 && t != 1 && t != n-1) {
    25         t = mul(t,t,n);
    26         d <<= 1;
    27     }
    28     return (d&1) || t == n-1;
    29 }
    30 bool isP(LL n) {
    31     if(n == 2) return true;
    32     if(n < 2 || 0 == (n&1)) return false;
    33     static int p[5] = {2,3,7,61,24251};
    34     for(int i = 0; i < 5; ++i)
    35         if(!check(p[i],n-1,n)) return false;
    36     return true;
    37 }
    38 LL gcd(LL a,LL b) {
    39     if(a < 0) return gcd(-a,b);//特别注意,没这个TLE
    40     return b?gcd(b,a%b):a;
    41 }
    42 LL Pollard_rho(LL n,LL c) {
    43     LL i = 1,k = 2,x = rand()%n,y = x;
    44     while(true) {
    45         x = (mul(x,x,n) + c)%n;
    46         LL d = gcd(y - x,n);
    47         if(d != 1 && d != n) return d;
    48         if(y == x) return n;
    49         if(++i == k) {
    50             y = x;
    51             k <<= 1;
    52         }
    53     }
    54 }
    55 LL Fac[maxn],tot;
    56 void factorization(LL n) {
    57     if(isP(n)) {
    58         Fac[tot++] = n;
    59         return;
    60     }
    61     LL p = n;
    62     while(p >= n) p = Pollard_rho(p,rand()%(n-1)+1);
    63     factorization(p);
    64     factorization(n/p);
    65 }
    66 int main() {
    67     int kase;
    68     LL x;
    69     scanf("%d",&kase);
    70     while(kase--) {
    71         scanf("%I64d",&x);
    72         if(isP(x)) puts("Prime");
    73         else {
    74             tot = 0;
    75             LL ret = x;
    76             factorization(x);
    77             for(int i = 0; i < tot; ++i)
    78                 if(ret > Fac[i]) ret = Fac[i];
    79             printf("%I64d
    ",ret);
    80         }
    81     }
    82     return 0;
    83 }
    View Code
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  • 原文地址:https://www.cnblogs.com/crackpotisback/p/4741916.html
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